//Given an array of integers numbers that is already sorted in ascending order, 
//find two numbers such that they add up to a specific target number. 
//
// Return the indices of the two numbers (1-indexed) as an integer array answer 
//of size 2, where 1 <= answer[0] < answer[1] <= numbers.length. 
//
// You may assume that each input would have exactly one solution and you may no
//t use the same element twice. 
//
// 
// Example 1: 
//
// 
//Input: numbers = [2,7,11,15], target = 9
//Output: [1,2]
//Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
// 
//
// Example 2: 
//
// 
//Input: numbers = [2,3,4], target = 6
//Output: [1,3]
// 
//
// Example 3: 
//
// 
//Input: numbers = [-1,0], target = -1
//Output: [1,2]
// 
//
// 
// Constraints: 
//
// 
// 2 <= numbers.length <= 3 * 104 
// -1000 <= numbers[i] <= 1000 
// numbers is sorted in increasing order. 
// -1000 <= target <= 1000 
// Only one valid answer exists. 
// 
// Related Topics 数组 双指针 二分查找 
// 👍 515 👎 0


package leetcode.editor.cn;

//Java：Two Sum II - Input array is sorted
class P167TwoSumIiInputArrayIsSorted {
    public static void main(String[] args) {
        Solution solution = new P167TwoSumIiInputArrayIsSorted().new Solution();
        // TO TEST
        solution.twoSum(new int[]{2, 7, 11, 15}, 9);
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        //1.双指针
        public int[] twoSum(int[] numbers, int target) {
            int left = 0, right = numbers.length - 1;
            while (left <= right) {
                int sum = numbers[left] + numbers[right];
                if (sum == target) {
                    return new int[]{left + 1, right + 1};
                }
                if (sum < target) {
                    left++;
                } else {
                    right--;
                }
            }
            return null;
        }

        //2.二分
        public int[] twoSumDiv(int[] numbers, int target) {
            int start = 0;
            while (start < numbers.length - 1) {
                int next = target - numbers[start];
                int left = start + 1, right = numbers.length - 1;
                while (left <= right) {
                    int mid = left + (right - left) / 2;
                    if (numbers[mid] == next) {
                        return new int[]{start + 1, mid + 1};
                    }
                    if (numbers[mid] < next) {
                        left = mid + 1;
                    } else {
                        right = mid - 1;
                    }
                }
                start++;
            }

            return null;
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}